Consider a 3 ×3matrix A whose (i,j)^-th element, a_i,j = (i-j)³ . Then the matrix A will be

The correct answer is skew-symmetric.

To determine the nature of the matrix $A$, we need to examine the relation between its elements $a_{i,j}$ and their transposed counterparts $a_{j,i}$.

The elements of the $3\times 3$ matrix $A$ are given by the formula:
$a_{i,j}={(i-j)}^3$
for all indices $i$ and $j$.

Now, let us find the expression for the transposed element $a_{j,i}$ by swapping the row index $i$ and the column index $j$:
$a_{j,i}={(j-i)}^3$
We can factor out $-1$ from the term inside the parenthesis:
$j-i=-(i-j)$
Substituting this back into the expression for $a_{j,i}$, we get:
$a_{j,i}={[-(i-j\left)\right]}^3$
Since the exponent $3$ is an odd integer, we have ${(-1)}^3=-1$. Therefore:
$a_{j,i}=-{(i-j)}^3$
Comparing this with our original formula for $a_{i,j}$, we have:
$a_{j,i}=-a_{i,j}$
for all $i$ and $j$.

By definition, a square matrix $A$ is skew-symmetric if its transpose satisfies $A^T=-A$, which is equivalent to the element-wise condition:
$a_{j,i}=-a_{i,j}$
Since this condition holds true for all elements of the matrix, the matrix $A$ is a skew-symmetric matrix.